Here, y = -15 with x = -20
LCM of 15 & 20 is 60, so divide 15 & 20 from 60,
60/15 = 4 & 60/20 = 3
Now, multiply the results with intial values,
4y = 3x
y = 3/4 x
When y = 12,
x = 12 * 4/3
x = 48 / 3
x = 16
In short, Your Answer would be 16
Hope this helps!
= 19 this is the answer I'm guessing since 16-12= 4 so 16 is 4 more than 12
Take any 2 points through which the line on graph passes,
(0,-1) =(x₁,y₁)
& (2,-4)=(x₂,y₂)
slope=(y₂-y₁)/(x₂-x₁)
=[-4-(-1)]/(2-0)
=-3/2
Answer:
wait is it y=20x + 40y = 30x+ 10
or y= 20x+ 40 y= 30x + 10
I cant really help you otherwise sorry when you clarify it i will edit the correct answer in
Answer:
The approximate length of the garden is at least 22 feet and at most 51 feet.
Step-by-step explanation:
A rectangular garden must have a perimeter of 145 feet.
Let x feet be the lengths of the garden. Then the width of the garden can be found using the perimeter:
The area of the garden is
The area must be at least 1,100 square feet, then
Solve this quadratic inequality:
![D=(-72.5)^2-4\cdot 1,100=856.25\\ \\\sqrt{D}=25\sqrt{1.37}\\ \\x_1=\dfrac{72.5-25\sqrt{1.37}}{2}\approx 22\\ \\x_2=\dfrac{72.5+25\sqrt{1.37}}{2}\approx 51\\ \\(x-22)(x-51)\le 0\\ \\x\in [22,51]](https://tex.z-dn.net/?f=D%3D%28-72.5%29%5E2-4%5Ccdot%201%2C100%3D856.25%5C%5C%20%5C%5C%5Csqrt%7BD%7D%3D25%5Csqrt%7B1.37%7D%5C%5C%20%5C%5Cx_1%3D%5Cdfrac%7B72.5-25%5Csqrt%7B1.37%7D%7D%7B2%7D%5Capprox%2022%5C%5C%20%5C%5Cx_2%3D%5Cdfrac%7B72.5%2B25%5Csqrt%7B1.37%7D%7D%7B2%7D%5Capprox%2051%5C%5C%20%5C%5C%28x-22%29%28x-51%29%5Cle%200%5C%5C%20%5C%5Cx%5Cin%20%5B22%2C51%5D)
